Lecture_4

Lecture_4 - Ch 3 Kinematics in 2D Now let’s consider the...

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Unformatted text preview: Ch. 3 Kinematics in 2D Now let’s consider the concepts of displacement , velocity , and acceleration in 2 dimensions. o f x x x- = Δ t x x t x v o f- = Δ Δ = t v v t v a o f- = Δ Δ = 2 2 1 at t v x x o o f + + = a v v x o f 2 2 2- = Δ 1-D o f r r r v v v- = Δ t r r t r v o f v v v v- = Δ Δ = t v v t v a o f v v v v- = Δ Δ = 2 2 1 t a t v r r o o f v v v v + + = a v v r o f v v v v 2 2 2- = Δ 2-D In general, for 2D, displacements, velocities, and accelerations will have components in both the x- and y-direction. Concepts from Ch. 1 Thus, y x r r r v v v + = y x v v v v v v + = y x a a a v v v + = 2 2 1 t a t v x x x o o f x + + = 2 2 1 t a t v y y y o o f y + + = We get 2 equations of motion under constant acceleration – one for the x- direction, and one for the y-direction. Example : A car drives 60 o N of E at a constant speed of 35 m/s. How far east has the car traveled after 10 s? E, x N, y v = 35 m/s 60 o We can break the velocity down into its x- and y-components. v y v x Now I can calculate the values for v x and v y : o x v v 60 cos = o y v v 60 sin = m/s 5 . 17 ) m/s)( 35 ( 2 1 = = m/s 3 . 30 ) m/s)( 35 ( 2 3 = = For distance in the x-direction (east) we use v x : t v x x = Δ m 175 s) m/s)(10 17.5 ( = = 3.3 Projectile Motion Now let’s analyze the motion of projectiles launched into the air, but this time, the motion is not completely vertical – 2-D projectile motion....
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Lecture_4 - Ch 3 Kinematics in 2D Now let’s consider the...

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